Menu Close

Author: Tinku Tara

Question-143024

Question Number 143024 by Feruzbek last updated on 09/Jun/21 Answered by bramlexs22 last updated on 09/Jun/21 $$\left(\mathrm{1}\right)\Leftrightarrow\:\mathrm{49}−{x}^{\mathrm{2}} \:\leqslant\:\mathrm{24}\: \\ $$$$\Leftrightarrow\:\mathrm{25}\:−{x}^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Leftrightarrow\:\left({x}+\mathrm{5}\right)\left({x}−\mathrm{5}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:{x}\leqslant−\mathrm{5}\:\cup\:{x}\geqslant\mathrm{5}…

Question-143027

Question Number 143027 by mathlove last updated on 09/Jun/21 Commented by Rasheed.Sindhi last updated on 09/Jun/21 $${x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1}\Rightarrow{x}=\omega,\omega^{\mathrm{2}} \\ $$$${When}\:{x}=\omega \\ $$$${x}^{\mathrm{123456789}} +\frac{\mathrm{1}}{{x}^{\mathrm{123456789}} } \\ $$$$\:\:\:\:\:=\omega^{\mathrm{123456789}}…